Figure 5 is a velocity/time graph for a lift moving upwards in a tall building - Edexcel - GCSE Physics Combined Science - Question 3 - 2023 - Paper 1

To find the total distance traveled by the lift, we will apply the formula for distance based on velocity and time:

$ext{Distance} = ext{Velocity} imes ext{Time}$

The lift has primarily three segments of movement:

1. From 0 to 6 seconds, the lift moves at 5 m/s for 6 seconds:

   $ext{Distance}_1 = 5 ext{ m/s} imes 6 ext{ s} = 30 ext{ m}$

2. From 6 to 12 seconds, the lift moves at 0 m/s (stationary) for 6 seconds:

   $ext{Distance}_2 = 0 ext{ m/s} imes 6 ext{ s} = 0 ext{ m}$

3. From 12 to 18 seconds, the lift again moves at 5 m/s for another 6 seconds:

   $ext{Distance}_3 = 5 ext{ m/s} imes 6 ext{ s} = 30 ext{ m}$

4. From 18 to 24 seconds, the lift moves downward at -5 m/s for 6 seconds:

   $ext{Distance}_4 = -5 ext{ m/s} imes 6 ext{ s} = -30 ext{ m}$

Now, add the distances together to calculate the total distance:

$ext{Total Distance} = ext{Distance}_1 + ext{Distance}_2 + ext{Distance}_3 + ext{Distance}_4$

Substituting the values gives:

$ext{Total Distance} = 30 ext{ m} + 0 ext{ m} + 30 ext{ m} - 30 ext{ m} = 30 ext{ m}$

Thus, the total distance traveled by the lift in 24 seconds is 30 meters.